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d of f over K(u1, · · · , un) is just N by definition. From the symmetric rational function theorem, we know that the splitting field of g over F is just L, and So N ∼= L. So we have an isomorphism Gal(N/K(u1, · · · , un)) → Gal(L/F ) ∼= Sn. Since Sn is not soluble, f is not soluble. This is our second main goal of the... |
ing the coefficients of f mod p. We also assume this has no repeated roots, and let ¯E be the splitting field of ¯f . Then there is an injective homomorphism ¯G = Gal( ¯E/Fp) → G = Gal(E/Q). Moreover, if ¯f factors as a product of irreducibles of length n1, n2, · · · , nr, then Gal(f ) contains an element of cycle type... |
ween HomK(K(α), E) ←→ RootPα (E) = {α1, · · · , αd}. wlog we let α = α1. Also, since we get | HomK(L, E)| = [L : K], | HomK(K(α), E)| = [K(α) : K] = deg Pα. 67 4 Computational techniques II Galois Theory Moreover, the restriction map HomK(L, E) → HomK(K(α), E) (defined by φ → φ|K(α)) is surjective and sends exactly [K(... |
ic triangles are rather nice. They are so nice that we can actually prove something about them! Proposition. For every geodesic triangulation of S2 (and respectively T ) has e = 2 (respectively, e = 0). Of course, we know this is true for any triangulation, but it is difficult to prove that without algebraic topology. Pr... |
write this as 1 0 ˙γ, ˙γ 1 2 γ(t) dt. Definition (Area). The area of a region W ⊆ V is defined as W (EG − F 2) 1 2 du dv when this integral exists. In the area formula, what we are integrating is just the determinant of the metric. This is also known as the Gram determinant. We define the distance between two points P and... |
olic line through z1, z2 (parametrized monotonically). Thus PSL(2, R) preserves hyperbolic distances. Similar to Euclidean space and the sphere, we show these lines minimize distance. Proposition. If γ : [0, 1] → H is a piecewise C 1-smooth curve with γ(0) = z1, γ(1) = z2, then length(γ) ≥ ρ(z1, z2), with equality iff γ... |
ay to view the hyperbolic plane as a subset of R3, and hence we need to mess with Riemannian metrics. However, it turns out we can indeed embed the hyperbolic plane in R3, if we give R3 a different metric! Definition (Lorentzian inner product). The Lorentzian inner product on R3 has the matrix 1 41 4 Hyperbolic geometr... |
∈ (−ε, ε). γτ = h( · , τ ) : [a, b] → V Proposition. A smooth curve γ satisfies the geodesic ODEs if and only if γ is a stationary point of the energy function for all proper variation, i.e. if we define the function E(τ ) = energy(γτ ) : (−ε, ε) → R, then dE dτ τ =0 = 0. 48 5 Smooth embedded surfaces (in R3) IB Geometry... |
he form γ(t) = σ(u0, t) for fixed u0. Meridians are curves of the form γ(t) = σ(t, v0) for fixed v0. These are generalizations of the notions of longitude and latitude (in some order) on Earth. In a general surface of revolution, we can compute the first fundamental form with respect to σ as E = σu2 = f 2 + g2 = 1, F = σu... |
the Riemannian metric on Vi, while on the left, we are computing it on Vj. Then we can define lengths, areas, energies on an abstract surface S. It is clear that every embedded surface is an abstract surface, by forgetting that it is embedded in R3. Example. The three classical geometries are all abstract surfaces. (i) ... |
pth in the next chapter. In this chapter, we will first look at results for general cosets. In particular, we will, step by step, prove the things we casually claimed above. Definition (Cosets). Let H ≤ G and a ∈ G. Then the set aH = {ah : h ∈ H} is a left coset of H and Ha = {ha : h ∈ H} is a right coset of H. Example... |
f. conjugacy classes) From the example last time, H = ⟨s⟩ ≤ D6 is not a normal subgroup, but K = ⟨r⟩ ◁ D6. We know that every group G has at least two normal subgroups {e} and G. Lemma. (i) Every subgroup of index 2 is normal. (ii) Any subgroup of an abelian group is normal. Proof. (i) If K ≤ G has index 2, then there ... |
tion is faithful if the kernel is just {e}. 5.2 Orbits and Stabilizers Definition (Orbit of action). Given an action G on X, the orbit of an element x ∈ X is orb(x) = G(x) = {y ∈ X : (∃g ∈ G) g(x) = y}. Intuitively, it is the elements that x can possibly get mapped to. Definition (Stabilizer of action). The stabilizer ... |
An. So |G| ≤ n!/2 We have seen on Sheet 1 that if |G| is even, then G has an element of order 2. In fact, Theorem (Cauchy’s Theorem). Let G be a finite group and prime p dividing |G|. Then G has an element of order p (in fact there must be at least p − 1 elements of order p). It is important to remember that this only ... |
n(F )). GLn(F ) = {A ∈ Mn×n(F ) : A is invertible} is the general linear group. Alternatively, we can define GLn(F ) as matrices with non-zero determinants. Proposition. GLn(F ) is a group. Proof. Identity is I, which is in GLn(F ) by definition (I is its self-inverse). The composition of invertible matrices is inverti... |
metries Consider the reflection in the mid-point of the cube τ , sending every point to its opposite. We can view this as −I in R3. So it commutes with all other symmetries of the cube. Proposition. G ∼= S4 × C2, where G is the group of all symmetries of the cube. Proof. Let τ be “reflection in mid-point” as shown abov... |
it here is that the M¨obius map satisfies this property. Proposition. The M¨obius group M acts sharply three-transitively on C∞. Proof. We want to show that we can send any three points to any other three points. However, it is easier to show that we can send any three points to 0, 1, ∞. Suppose we want to send z1 → ∞,... |
erm is divisible by p. Also |G| = pn is divisible by p. Hence the number of conjugacy classes of size 1 is divisible by p. We know {e} is a conjugacy class of size 1. So there must be at least p conjugacy classes of size 1. Since the smallest prime number is 2, there is a conjugacy class {x} = {e}. But if {x} is a conj... |
larly, the largest power of p dividing pa − j is also the largest power of p dividing j. So we have the same power of p on top and 24 1 Groups IB Groups, Rings and Modules bottom for each item in the product, and they cancel. So the result is not divisible by p. This proof is not straightforward. We first needed the cle... |
ith the constant polynomials, i.e. polynomials aiX i with ai = 0 for i > 0. In particular, 0R ∈ R and 1R ∈ R are the zero and one of R[X]. This is in fact a ring. Note that a polynomial is just a sequence of numbers, interpreted as the coefficients of some formal symbols. While it does indeed induce a function in the obv... |
0. Similarly, if deg g = m, then g = m i=0 biX i, with bm = 0. If n < m, we let q = 0 and r = f , and done. Otherwise, suppose n ≥ m, and proceed by induction on n. We let f1 = f − anb−1 m X n−mg. This is possible since bm = 0, and F is a field. Then by construction, the coefficients of X n cancel out. So deg(f1) < n. If... |
{(a, b) ∈ R × R : b = 0}. We think of (a, b) ∈ S as a b . We define the equivalence relation ∼ on S by (a, b) ∼ (c, d) ⇔ ad = bc. We need to show this is indeed a equivalence relation. Symmetry and reflexivity are obvious. To show transitivity, suppose i.e. (a, b) ∼ (c, d), (c, d) ∼ (e, f ), ad = bc, cf = de. We multiply... |
by N : R → Z≥0 √ N (a + b −5) → a2 + 5b2. It is convenient to think of this as z → z ¯z = |z|2. This satisfies N (z · w) = N (z)N (w). This is a desirable thing to have for a ring, since it immediately implies all units have norm 1 — if r · s = 1, then 1 = N (1) = N (rs) = N (r)N (s). So N (r) = N (s) = 1. So to find th... |
tored as a product of irreducibles (if both can, then r would be a product of irreducibles). So we can write r1 = r2s2, with r2, s2 not units. Again, wlog r2 cannot be factored as a product of irreducibles. We continue this way. By assumption, the process does not end, and then we have the following chain of ideals: (r... |
product in R[X]. So we have So we have Cancelling b gives c(bf ) = c(b)c(f ) = bc(f ). bf = ubc(f )gh1. f = g(uc(f )h1). So g | f in R[X]. So f ∈ J. From this we can get ourselves a large class of UFDs. Theorem. If R is a UFD, then R[X] is a UFD. In particular, if R is a UFD, then R[X1, · · · , Xn] is also a UFD. Proof... |
. As the zi are not units, we know N (z1) = N (z2) = p. By definition, this means p = z1 ¯z1 = z2 ¯z2. But also p = z1z2. So we must have ¯z1 = z2. Finally, we have p = z1 ¯z1 | N (z) = z ¯z. All these z, zi are irreducible. So z must be an associate of z1 (or maybe ¯z1). So in particular N (z) = p. Corollary. An intege... |
ortant case we have not tackled yet — polynomial rings. We know Z[X] is not a PID, since (2, X) is not principal. However, this is finitely generated. So we are not dead. We might try to construct some non-finitely generated ideal, but we are bound to fail. This is since Z[X] is a Noetherian ring. This is a special case ... |
annihilator is a subset of R. Moreover it is an ideal — if r · m = 0 and s · m = 0, then (r + s. So r + s ∈ Ann(m). Moreover, if r · m = 0, then also (sr) · m = s · (r · m) = 0. So sr ∈ Ann(m). What is this good for? We first note that any m ∈ M generates a submodule Rm as follows: Definition (Submodule generated by elem... |
ring has a maximal ideal. This is a rather strong statement, since it talks about “all rings”, and we can have weird rings. We need to use a more subtle argument, namely via Zorn’s lemma. You probably haven’t seen it before, in which case you might want to skip the proof and just take the lecturer’s word on it. Proof. ... |
the effect of swapping rows or multiplying rows on determinants. So after any row operation, the resultant submatrix C satisfies det(C ) ∈ Fitk(A). Since this is true for all minors, we must have But row operations are invertible. So we must have Fitk(B) ⊆ Fitk(A). as well. So they must be equal. So done. Fitk(A) ⊆ Fitk(... |
or a field F, the polynomial ring F[X] is a Euclidean domain, so the results of the last few sections apply. If V is a vector space on F, and α : V → V is a linear map, then we can make V into an F[X]-module via F[X] × V → V (f, v) → (f (α))(v). We write Vα for this F[X]-module. Lemma. If V is a finite-dimensional vector... |
g at eigenspaces. So these cases are disjoint. Note that we have done more work that we really needed, since λI is invariant under conjugation. But the first case is not too satisfactory. We can further classify it as follows. If X 2 + a1X + a2 is reducible, then it is for some µ, λ ∈ F. If λ = µ, then the matrix is con... |
ll g 2 G. An intersection of normal subgroups of a group is again a normal subgroup (cf. 1.14). Therefore, we can define the normal subgroup generated by a subset X of a group G to be the intersection of the normal subgroups containing X. Its description in terms of X is a little complicated. We say that a subset X of ... |
1;:::;hk /7!.h1hk1;hk / ! H1 Hk1 Hk .h;hk /7!hhk ! G sends .h1; h2; : : : ; hk/ to h1h2 hk: Commutative groups 25 Commutative groups The classification of finitely generated commutative groups is most naturally studied as part of the theory of modules over a principal ideal domain, but, for the sake of completeness, I ... |
up F having orders 2m, 2n, and 2r respectively. Let q a D u 1 0 u1 and b D v t 0 v1 (elements of SL2.Fq/); where t has been chosen so that uv C t C u1v1 D w C w1: The characteristic polynomial of a is .X u/.X u1/, and so a is similar to diag.u; u1/. Therefore a has order 2m. Similarly b has order 2n. The matrix ab D uv... |
ts a unique homomorphism ˛W FX ! F 0 such that ˛ ı D 0; by the universality of 0, there exists a unique homomorphism ˇW F 0 ! FX such that ˇ ı 0 D ; now .ˇ ı ˛/ ı D , but by the universality of , idF X is the unique homomorphism FX ! FX such that idF X ı D , and so ˇ ı ˛ D idF X ; similarly, ˛ ı ˇ D idF 0, and so ˛ and... |
See Feit 1995. TODD-COXETER ALGORITHM There are some quite innocuous looking finite presentations that are known to define quite small groups, but for which this is very difficult to prove. The standard approach to these questions is to use the Todd-Coxeter algorithm (see Chapter 4 below). We shall develop various met... |
Let ˛ be an automorphism of a group H . If ˛ is inner, then it extends to every group G containing H as a subgroup. The converse is also true (Schupp 1987). COMPLETE GROUPS DEFINITION 3.3 A group G is complete if the map g 7! ig W G ! Aut.G/ is an isomorphism. Thus, a group G is complete if and only if (a) the centre Z... |
inner. Let ˛ be an automorphism, possibly outer, of a group N . We can realize N as a normal subgroup of a group G in such a way that ˛ becomes the restriction to N of an inner automorphism of G. To see this, let W C1 ! Aut.N / be the homomorphism sending a generator a of C1 to ˛ 2 Aut.N /, and let G D N Ì C1. The elem... |
termination of the finite simple groups with an involution centralizer isomorphic to H is a finite problem. Later work showed that the problem is even tractable, and so the strategy became: (a) list the groups H that are candidates for being an involution centralizer in some finite simple group, and (b) for each H in (... |
section \ CG.x/ D fg 2 G j gx D xg for all x 2 Gg x2G 60 4. GROUPS ACTING ON SETS is the centre of G. (b) Let G act on G=H by left multiplication. Then Stab.H / D H , and the stabilizer of gH is gHg1: (c) Let G be the group of rigid motions of Rn (4.2f). The stabilizer of the origin is the orthogonal group On for the s... |
a) H is normal in G, or (b) H is not normal in G. In the first case, rsr 1 D s, i.e., rs D sr, and so G ' hri hsi C2 C3. In the second case, G ! Sym.G=H / is injective, hence surjective, and so G S3 D3. Permutation groups Consider Sym.X/, where X has n elements. Since (up to isomorphism) a symmetry group Sym.X/ depends... |
4.36 Every normal subgroup N of An, n 5, N ¤ 1, contains a cycle of length 3. PROOF. Let 2 N , ¤ 1. If is not a 3-cycle, we shall construct another element 0 2 N , 0 ¤ 1, which fixes more elements of f1; 2; : : : ; ng than does . If 0 is not a 3-cycle, then we can apply the same construction. After a finite number of s... |
t of lines through the origin in F3 2; show that X has 7 elements, and that there is a natural injective homomorphism G ,! Sym.X/ D S7. (c) Use Jordan canonical forms to show that G has six conjugacy classes, with 1, 21, 42, 56, 24, and 24 elements respectively. [Note that if M is a free F2Œ˛-module of rank one, then E... |
ich contradicts what we proved in the last paragraph. There can be no such P , and so O is all of X. (b) Since sp is now the number of elements in O, we have also shown that sp 1 (mod p/. Let P be a Sylow p-subgroup of G. According to (a), sp is the number of conjugates of P , which equals .G W NG.P // D .G W 1/ .NG.P ... |
xpW G ! G is a homomorphism. Its image is contained in Z.G/, and so its kernel has order at least p2. Since N contains only p 1 elements of order p, we see that there exists an element b of order p outside N . Hence G D hai Ì hbi Cp2 Ì Cp, and it remains to observe (3.19) that the nontrivial homomorphisms Cp ! Aut.Cp2/... |
er the subgroups B D of GL2.F /, some field F . Then U is a normal subgroup of B, and B=U ' F F , U ' .F; C/. Hence B is solvable. and U D 0 1 0 1 PROPOSITION 6.6 (a) Every subgroup and every quotient group of a solvable group is solvable. (b) An extension of solvable groups is solvable. PROOF. (a) Let G F G1 F F Gn be... |
on follows from the preceding corollary. For the converse, let G be a finite nilpotent group. According to (5.9) it suffices to prove that all Sylow subgroups are normal. Let P be such a subgroup of G, and let N D NG.P /. The first lemma below shows that NG.N / D N , and the second then implies that N D G, i.e., that P... |
a ring A containing F in its centre and finite dimensional as an F -vector space. We do not assume A to be commutative; for example, A could be the matrix algebra Mn.F /. Let fe1; : : : ; eng be a basis for A as an F -vector space; then ei ej D P ij ekfor 2 F , called the structure constants of A relative to the basis;... |
g g; cg 2 F; g 2 G, two elements of F ŒG are equal, X g cg g D X c0 g g, g if and only if cg D c0 g for all g, and X g cg g X g c0 g g D X g c00 g g; with c00 g D P g1g2Dg cg1c0 g2. A linear action g; v 7! gvW G V ! V of G on an F -vector space extends uniquely to an action of F ŒG on V , P g cg g; v 7! P g cg gvW F ŒG... |
av1 D bv1; av2 D bv2; : : : ; avn D bvn: PROOF. We first prove this for n D 1. Note that Av1 is an A-submodule of V , and so (see 7.13) there exists an A-submodule W of V such that V D Av1 ˚ W . Let W V ! V be the map .av1; w/ 7! .av1; 0/ (projection onto Av1). It is A-linear, hence lies in D, and has the property that... |
ices to show that they span the centre. For any g 2 G and P a2G maa 2 F ŒG, X maa a2G g g1 D X magag1: a2G The coefficient of a in the right hand sum is mg 1ag , and so g X a2G g1 D X maa mg 1ag a: a2G This shows that P constant on conjugacy classes, i.e., if and only if P a2G maa lies in the centre of F ŒG if and only... |
˛.f / D gf for all f 2 A. [Hint: Deduce from (b) that there exists a bijection W G ! G such that .˛f / .g/ D f .g/ for all g 2 G. From the hypothesis on ˛, deduce that .g1g2/ D g1 .g2/ for all g1; g2 2 G.R/. Hence .g/ D g .e/ for all g 2 G. Deduce that ˛.f / D .e/f for all f 2 A.] (f) Show that the following maps are G... |
ed subgroup. Prove that L D K. [You may assume that a finite group is nilpotent if and only if every maximal subgroup is normal.] (b) Let G be a finite group. If G has a subgroup H such that both G=ıH and H are nilpotent, prove that G is nilpotent. 76. Let G be a finite noncyclic p-group. Prove that the following are e... |
; 1/, and .1; : : : ; 1/ has order p in .Cp/p. (e) We have .n; q/.n; q/ D .nn1; qq/ D .1; 1/: 3-8 Let n q 2 Z.G/. Then .n q/.1 q0/ D .1 q0/.n q/ D q0nq01 q0q n qq0 all q0 2 Q H) n 2 CN .Q/ q 2 Z.Q/ and .n q/.n0 1/ D nq n0q1 q .n0 1/.n q/ D n0n q n0 2 N H) n1n0n D q n0q1: The converse and the remaining statements are ea... |
b) It’s the quaternion group. From the two relations get yx D x1y; yx D xy1 and so x2 D y2. The second relation implies xy2x1 D y2; D y2; and so y4 D 1. Alternatively, the Todd-Coxeter algorithm shows that it is the subgroup of S8 generated by .1287/.3465/ and .1584/.2673/. Bibliography ALPERIN, J. L. AND BELL, R. B. 1... |
hmetic progressions, with fixed 𝑘 𝑐𝑘 (Gowers 2001). /( As for lower bounds, Behrend (1946) constructed a subset of 𝑐√log 𝑁 that avoids 3-term arithmetic progressions. This is an important construction that we will see in Section 2.5. Some researchers think that this lower bound is closer to the truth, since for a ... |
to 𝑥 – Proof: set up a graph whose triangles correspond to solutions to 𝑥 𝑦 = 𝑧, and then 𝑦 = 𝑧. + + apply Ramsey’s theorem. Szemerédi’s theorem. Every subset of N with positive density contains arbitrarily long arithmetic progressions. – A foundational result that led to important developments in additive combin... |
-free graph is empty. Now assume 𝑟 > 1 and that ex 𝑛, 𝐾𝑟 ) 1-free graph. Let 𝑣 be a vertex of maximum degree in 𝐺. Since 𝐺 for every 𝑛. Let 𝐺 = 𝑇𝑛,𝑟 = 𝑒 1) ( ( − 𝑉, 𝐸 be a 𝐾𝑟 1-free, the neighborhood 𝐴 = 𝑁 ) ( + is 𝐾𝑟 + 𝑣 of 𝑣 is 𝐾𝑟 -free. So, by the induction hypothesis, ) ( ex 𝐴 𝑒 ( ) ≤ 𝐴 ... |
niform hypergraph 𝐻, we write ex for the maximum number of edges in an 𝑛-vertex 𝑟-uniform hypergraph that does not contain 𝐻 as a subgraph. A straightforward 𝑛 Graph Theory and Additive Combinatorics — Yufei Zhao 1.3 Turán Density and Supersaturation 21 extension of Proposition 1.3.1 gives that ex each fixed 𝐻. S... |
that by choosing the optimal 𝑟 as a function of 𝑛, we can get at least 𝑐 𝑛1 + / log log 𝑛 unit distances, where 𝑐 > 0 is some absolute constant. The proof uses analytic number theory, which we omit as it would take us too far afield. The basic idea is to choose 𝑟 to be a product of many distinct primes that are... |
ce 𝑚 = 𝑂 𝑠 ( )𝑟 ,𝑠,𝑡 ) 3 1 𝑛3 − / 𝑠2 . ) 𝑛3 □ 𝑛, 𝐾 ( Exercise 1.5.8. Prove that ex = 𝑂𝑟 ,𝑠,𝑡 ( We can iterate further, using the same technique, to prove an analogous result for every uniformity, thereby giving us the statement (Theorem 1.5.6) used in our proof of the Erdős–Stone–Simonovits theorem earli... |
e expected to share some common friends – at least they all know Alice. We can take a step further, and pick a few people at random (Alice, Bob, Carol, David) and have them host a party and invite all their common friends. This will likely be an even more sociable crowd. At least all the party goers will know all the h... |
𝑛, 𝐾4,4) In general, given 𝐻, we should apply Theorem 1.9.1 to the subgraph of 𝐻 with the ratio. This gives the following corollary, which sometimes maximum 2 ) gives a better lower bound than directly applying Theorem 1.9.1. ) ≥ )/( ) − ) − ex 𝐻 𝐻 1 𝑒 𝑣 ( ( ( ( ( − / ≳ 𝑛2 6 / − 15. − / Definition 1.9.2 (2-den... |
𝑎11)( 𝑎21)( 𝑥2 − 𝑥2 − 𝑎12) · · · ( 𝑎22) · · · ( 𝑥𝑠 − 𝑥𝑠 − 𝑎1𝑠) 𝑎2𝑠) = 𝑏1 = 𝑏2 ... 𝑥1 − 𝑎𝑠2) · · · ( 𝑥2 − 𝑎𝑠1)( ( F𝑠. 𝑥1, . . . , 𝑥𝑠) ∈ ( 𝑥𝑠 − 𝑎𝑠𝑠) = 𝑏𝑠 has at most 𝑠! solutions 𝑖 ( ) ] [ 𝑠 𝑠 ] → [ , setting 𝑥𝑖 = 𝑎𝑖 𝜋 Remark 1.10.12 (Special case 𝑏 = 0). Consider the special c... |
ch value with ) □ probability 1 is uniformly distributed in F𝑞 for a fixed . Hence 𝑓 𝑢, 𝑣 𝑢, 𝑣 𝑢, 𝑣 𝑞. ) ) ( ( ( / More generally, we show below that the expected occurrence of small subgraphs mirrors for the set of unordered that of the usual random graph with independent edges. We write pairs of element from... |
the graph regularity method. Similar to the proof of Schur’s theorem in Chapter 0, this graph theoretic proof of Roth’s theorem demonstrates a fruitful connection between graph theory and additive combinatorics. By the regularity method, we mean both the graph regularity lemma as well as methods for applying it. Rather... |
en there exists a refinement 1 parts, and such that + of P Q Proof. Let 𝑞 (Q) > 𝑞 (P) + 𝜀5. 𝑅 = 𝑖, 𝑗 {( 𝑉𝑖, 𝑉 𝑗) 𝑘 2 : 𝑉𝑖, 𝑉 𝑗) is 𝜀-regular and 𝑅 = 2 𝑘 𝑅. ( ] ) ∈ [ that is not 𝜀-regular, find a pair 𝐴𝑖, 𝑗 \ 𝑉 𝑗 that witnesses For each pair ⊆ 𝑅. Note that for 𝑖 ≠ 𝑗, we can take the irregula... |
∗ (Regularity partition into regular sets). Show that for every 𝜀 > 0 there exists 𝑀 so that every graph has an 𝜀-regular partition into at most 𝑀 parts, with every part being 𝜀-regular with itself. 2.2 Triangle Counting Lemma Szemerédi’s regularity lemma gave us a vertex partition of a graph. How can we use this ... |
proof of Schur’s theorem in Chapter 0. We write 3-AP for “3-term arithmetic progression.” We say that 𝐴 is 3-AP-free if there are no 𝑥, 𝑥 𝑦, 𝑥 2𝑦 + ∈ + 𝐴 with 𝑦 ≠ 0. Theorem 2.4.1 (Roth’s theorem) Let 𝐴 be 3-AP-free. Then 𝑁 ⊆ [ ] 𝐴 | = 𝑜 𝑁 . ) ( | Proof. Embed 𝐴 / in Z, it is 3-AP-free in Z 𝑀Z with 𝑀 =... |
𝑦, 𝑥, 𝑧 𝐴 𝐴 not all equal and satisfying 𝑥 𝑁 exp 𝑦 log 𝑁 with ⊆ [ 𝑁 ] ) | 𝐶 (− 𝑧 = 3𝑤. √︁ + | ≥ + ∈ Exercise 2.5.5∗ (Avoiding 5-term quadratic configurations). Prove that there is some constant 𝐶 > 0 so that for all 𝑁, there exists 𝐴 𝑁 exp so that there does not exist a nonconstant quadratic polynomial... |
) is < 𝑚𝜀𝑛2 / ) ≤ 8 / 𝜀 4 𝜀𝑛2 ≤ 3 1 / ) 𝜒 ≤ ( 1 + ) − − ( By Turán’s theorem (Corollary 1.2.6), 𝐺 ′ contains a copy of 𝐾𝜒 𝐻 ) ) land in 𝑉𝑖1, . . . , 𝑉𝑖𝜒 vertices of this 𝐾𝜒 (allowing repeated indices). Since each pair of 𝐻 ( ) these sets is 𝜂-regular, has edge density 𝜀 , by applying / the graph co... |
m 2.8.9 (Strong regularity lemma) For any sequences of constants 𝜀0 ≥ 𝜀1 ≥ so that every 𝑛-vertex graph has an equitable vertex partition 𝑉1 ∪ · · · ∪ 𝑉𝑖 for each 𝑖 satisfying 𝑊𝑖 ⊆ 𝑊𝑖 | ≥ 𝛿𝑛, (a) | 𝑊𝑖, 𝑊 𝑗) (b) ( 𝑉𝑖, 𝑉 𝑗) − 𝑑 (c) ( is 𝜀𝑘-regular for all 1 𝑊𝑖, 𝑊 𝑗) 𝑗 ≤ 𝜀0 for all but < 𝜀0�... |
presenting edges and nonedges respectively. 𝐻𝑏𝑏𝑐𝑎𝑐−→𝜙𝑇𝑎𝑏𝑐blackgray(none)whiteGraph Theory and Additive Combinatorics — Yufei Zhao 84 Graph Regularity Method 2.9 Graph Property Testing We are given random query access to a very large graph. The graph may be too large for us to see every vertex or edge. What c... |
tially more difficult to prove than graph regularity. We only sketch some key ideas here. For concreteness, we focus our discussion on 3-graphs. Throughout this section, 𝐺 will be a 3-graph with vertex set 𝑉. What should correspond to an “𝜀-regular pair” from the graph regularity lemma? Here is an initial attempt. D... |
irandom Graphs Here are several natural notions of how a graph (or rather, a sequence of graphs) can look random. The main theorem of this section says that, surprisingly, these notions are all equivalent. This result is due to Chung, Graham, and Wilson (1989), who coined the term quasirandom graphs. Similar ideas also... |
thod in probabilistic combinatorics. The condition C4 says that the variance of the codegree of two random vertices is small. Exercise 3.1.17. Show that if we modify the CODEG condition to 𝑉 ∑︁𝑢,𝑣 ( ∈ then it would not be enough to imply quasirandomness. ) 𝐺 codeg 𝑢, 𝑣 ( ) − 𝑝2𝑛 = 𝑜 𝑛3 , ) ( Proof that CODEG ... |
m bipartite graphs are bipartite quasirandom. The proof (omitted) is essentially the same as Proposition 3.1.8 and Corollary 3.1.9. Proposition 3.1.28 (Random bipartite graphs are typically quasirandom) 𝑛, 𝑛, 𝑝 . With probability 1, a sequence of bipartite random graphs 𝐺 𝑛 ∼ Fix 𝑝 ) (obtained by keeping every ed... |
on and Milman (1985) and Alon (1986). | | ≤ | | / | | Theorem 3.2.13 (Cheeger’s inequality) Let 𝐺 be an 𝑛-vertex 𝑑-regular graph with adjacency matrix spectral gap 𝜅 = 𝑑 Then its edge-expansion ratio ℎ = ℎ satisfies 𝐺 𝜆2. − ( 𝜅 ) 2 ℎ √2𝑑𝜅. / ≤ ≤ The two bounds of Cheeger’s inequality are tight up to constant ... |
ondition EIG from Theorem 3.1.1. Theorem 3.3.12 (Eigenvalues of Paley graphs) Let 𝑝 ≡ top eigenvalue be a prime. The adjacency of matrix of the Paley graph of order 𝑝 has 2. 1 2, and all other eigenvalues are either mod 4 𝑝 2 or 1 1 1 ( √𝑝 √𝑝 ( − )/ (− − )/ ) − ( )/ Proof. Applying Theorem 3.3.8, we see that the e... |
riant, its C-span 𝑊 is a Γ-invariant subspace (i.e., Γ), and hence a subrepresentation of Γ. Since 𝑣 is not a constant vector, 𝑔𝑊 𝐾 the Γ-action on 𝑣 is nontrivial. So 𝑊 is a nontrivial representation of Γ. Hence dim 𝑊 by hypothesis. Every nonzero vector in 𝑊 is an eigenvector of 𝐴 with eigenvalue 𝜇. It foll... |
s is an advanced remark and can be skipped over.) Section 3.3 discussed the Fourier transform on finite abelian groups. The Graph Theory and Additive Combinatorics — Yufei Zhao 118 Pseudorandom Graphs topic of this section can be alternatively viewed through the lenses of the nonabelian Fourier transform. We refer to W... |
𝑑-regular tree. The following result gives the lower bound on 𝜆 (Alon 1986). Theorem 3.6.2 (Alon–Boppana second eigenvalue bound) Fix a positive integer 𝑑. Let 𝐺 be an 𝑛-vertex 𝑑-regular graph. If 𝜆1 ≥ · · · ≥ eigenvalues of its adjacency matrix, then 𝜆𝑛 are the where 𝑜 1 ( ) → 0 as 𝑛 . → ∞ 2√𝑑 𝜆2 ≥ 1 𝑜 ... |
graph is Ramanujan with probability at least 𝑐𝑑. ≥ If this were true, it would prove Conjecture 3.6.9 on the existence of Ramanujan graphs. However, no rigorous results are known in this vein. One can formulate a bipartite analog. Definition 3.6.12 (Bipartite Ramanujan graph) A bipartite Ramanujan graph is some bipar... |
icalities). We try to sweep measure theoretic technicalities under the rug in order to focus on key ideas. If you have not seen measure theory before, do not worry. Just view “measure” as lengths of intervals or areas of boxes (or countable unions thereof) in the most natural sense. We always ignore measure zero differ... |
: distance (or cut metric) to be 𝜹□(𝑼, 𝑾) inf 𝜙 𝑊 𝜙 𝑈 □ − 0, 1 2 ] [ → R, we define their cut sup 𝑆,𝑇 = inf 𝜙 ∫𝑆 × where the infimum is taken over all invertible measure preserving maps 𝜙 : 0, 1 [ associated graphons: ] → . Define the cut distance between two graphs 𝐺 and 𝐺 ′ by the cut distance of their... |
edges. R by 𝑊 𝐹, 𝑊 0, 1 𝑥, 𝑦 2𝜋 )) − .4 𝑊-Random Graphs In this section, we explain how to use a graphon to create a random graph model. This hopefully gives more intuition about graphons. The most common random graph model is the Erdős–Rényi random graph G is an 𝑛-vertex graph with every edge chosen with prob... |
om 𝑊 𝐹 starting with 𝑡𝐹 ( ( to 𝑈, one by one, resulting in a total change of at most 𝑒 □. This proves (4.6), □ and hence the theorem. 𝑊𝑒 : 𝑒 ) ∥ 𝑊 𝑊 𝑈 𝑈 )) − .6 Weak Regularity Lemma In Chapter 2, we defined an 𝜀-regular vertex partition of a graph to be a partition such that all but 𝜀-fraction of pairs ... |
that such that for every graphon 𝑊, there is a pair of partitions 𝜀1, 𝜀2, . . . 𝑀 (here refines 0, 1 and of Q| ≤ |Q | and ∥ 𝜀2 1. Furthermore, deduce the strong regularity lemma in the following form: 𝑊 = 𝑊str + 𝜀1. State your 𝑀, where 𝑊str is a 𝑘-step graphon with 𝑘 ≤ 𝑘 2 roughly bounds on 𝑀 explicitly ... |
sequence) → ∞ converges to 𝑈 in cut metric. Let 𝜀 > 0. Then there exists some 𝑘 > 3 𝜀 such that − convergence and the dominated convergence theorem. Then 𝛿□ there exists some 𝑛0 ∈ 𝑘 > 3 N such that 𝛿□ < 𝜀 / ∥ 𝑈 3, by pointwise 3. By (4.8), 3 for all 𝑛 > 𝑛0. Finally, since we chose 𝑈𝑘 ∥1 < 𝜀 𝑈, 𝑈𝑘) ( ... |
— Yufei Zhao 158 Graph Limits Similar to Definition 4.2.4 of the cut distance 𝛿□, define the distance based on the 𝐿1 norm: 𝜹1(𝑾, 𝑼) inf 𝜙 ∥ 𝑊 − 𝑈 𝜙 ∥1 . □ [ ≤ ∥ · ∥ 0, 1 0, 1 𝑘, 𝑊 ] → [ where the infimum is taken over all invertible measure preserving maps 𝜙 : 𝛿1. Since ≤ ∥ · ∥1, we have 𝛿□ Lemma 4.9.4 ... |
1). This means that there is no algorithm that always correctly decides whether a given inequality is true for all graphs (however, it does not prevent us from proving/disproving specific inequalities). This undecidability stands in stark contrast to the decidability of polynomial inequalities over the reals, which fol... |
x graph with 𝑚 edges? When 𝑚 = for some integer 𝑎, the optimal graph is a clique on 𝑎 vertices. More generally, for any value of 𝑚, the optimal graph is obtained by adding edges in colexicographic order: 𝑎 2 12, 13, 23, 14, 24, 34, 15, 25, 35, 45, . . . This is stronger than Theorem 5.1.5, which only gives an asy... |
g lower bound on the minimum triangle density given edge density (Goodman 1959). See Figure 5.2 for a plot. Theorem 5.2.8 (Lower bound on triangle density) 𝐾3, 𝑊 𝑡 ( ) ≥ 𝑡 ( 𝐾2, 𝑊 2𝑡 ( )( 𝐾2, 𝑊 1 . ) ) − 𝐾2, 𝑊 ( ) The inequality is tight whenever 𝑊 = 𝐾𝑛, in which case 𝑡 𝑛 . In particular, Goodman’s boun... |
1 1 / / 𝑓𝑘 ≤ ∥ where the 𝒑-norm of a function 𝑓 is defined by 𝑓1 𝑓2 · · · ∫ 𝑓1∥ 𝑝1 · · · ∥ 𝑓𝑘 ∥ 𝑝𝑘 , ∥ of Hölder’s inequality is used often. ∫ In practice, the case 𝑝1 = · · · We can apply Hölder’s inequality to show that 𝐾𝑠,𝑡 is Sidorenko. The proof is essentially 4 from the previous section, verbatim... |
𝑑2 . )/ ( Let us prove this theorem in the case 𝐹 = 𝐶6 to illustrate the technique more concretely. The general proof is basically the same. Let 𝑓 𝑥1, 𝑥2) ( = ∫𝑦 𝑊 ( 𝑥1, 𝑦 𝑊 ) ( 𝑥2, 𝑦 . ) This function should be thought of as the codegree of vertices 𝑥1 and 𝑥2. Then, grouping the factors in the integral... |
that be a real linear combination of elementary symmetric polynomials in R𝑛 with 𝑥 𝑥1, . . . , 𝑥𝑛) 𝑥𝑛 = 1, and furthermore 𝑥 has minimum support size among 𝑛 ( 0 and 𝑥1 + · · · + among all vectors 𝑥 minimizes 𝑓 ≤ ≤ ∈ 𝑘 ) ( 𝑥1 = · · · = 𝑥𝑘 = 1 and 𝑥𝑘 1 = + · · · = 𝑥𝑛 = 0. 𝑘 / 1 = 𝑥1, 𝑥2) = 𝑥𝑛 = ... |
, 𝑮) for the the set of all maps 𝑉 homomorphism 𝐹 random element Φ 𝐺. This set has cardinality hom Hom whose entropy satisfies 𝐹, 𝐺 ( 𝐹 ( 𝐹, 𝐺 𝑉 𝐺 that give a graph ) → . Our strategy is to construct ) log2( ) log2 hom ) ≥ Φ ( ) ≤ 2𝑒 𝐺 2𝑒 𝐹 𝑣 𝐹 ( 𝐹, 𝐺 )) − ( ( ) − )) then implies (5.5). ( ( ) The uni... |
h was in turn based on the proof by Kahn (2001) for independent sets. Proof. Let us first illustrate the proof for 𝐹 being the following graph uniformly at random among all homomorphisms from 𝐹 to 𝐺. Let 𝐺 be the respective images of the vertices of 𝐺. We have ) Hom 𝐹, 𝐺 Choose Φ ) ∈ 𝑉 𝑋1, 𝑋2, 𝑋3, 𝑌1, 𝑌2, ... |
𝜶∥ℓ2 𝛼, 𝛼 ⟨ 2 1 / ℓ2 . ⟩ F𝑛 𝑥 𝑝 ∑︁ ∈ Writing 𝛾𝑟 : F𝑛 𝑝 → C for the function defined by 𝜸𝒓 (𝒙) 𝜔𝑟 · 𝑥 (this is a character of the group F𝑛 ) Parseval’s identity can be stated as ( 𝑓 𝑟 = 𝑝), the Fourier transform can be written as = E𝑥𝛾𝑟 ( 𝛾𝑟 , 6.1ℓ2 ⟨ and 𝑓 ∥2 = ∥ ∥ 𝑓 ∥ℓ2 . With these conventi... |
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